Find the remainder when x³+3x²+3x+1 is divided by (i) x+1 (ii) x-1/2 (iii) x (iv) x+π (v) 5+2x

Using remainder theorem, if p(x) is divided by (x-a), then the remainder can be found by p(a)

Given: p(x) = x³+3x²+3x+1

(i) If x+1=0, then x=-1
So putting the value of x=-1 in the given equation, we get,
p(-1) = (-1)³+3(-1)²+3(-1)+1 = -1+3-3+1 = 0
Hence, remainder is 0

(ii) If x-1/2=0, then x=1/2
So, putting x=1/2 in equation, we get
p(1/2) = (1/2)³+3(1/2)²+3(1/2)+1 = 1/8+3/4+3/2+1 = 1/8+6/8+12/8+8/8 = 27/8 = 3 3/8
Hence, the remainder is 27/8

(iii) If x+0=0 then x=0
So putting the value of x=0 in the given equation, we get,
p(0) = (0)³+3(0)²+3(0)+1 = 0+0+0+1 = 1
Hence, the remainder is 1

(iv) If x+π=0, then x=-π
So putting the value of x=-π in the equation ,we get,
p(-π) = (-π)³+3(-π)²+3(-π)+1 = -π³+3π²-3π+1
Hence, the remainder is -π³+3π²-3π+1

(v) If 5+2x=0, then x=-5/2
So putting the value of x=-5/2 in the given equation, we get,
p(-5/2) = (-5/2)³+3(-5/2)²+3(-5/2)+1 = -125/8+75/4-15/2+1 = -125/8+150/8-60/8+8/8 = -27/8
Hence, the remainder is -27/8